/*
Missing Number
==============

Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.

Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?

Example 1:
Input: nums = [3,0,1]
Output: 2
Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.

Example 2:
Input: nums = [0,1]
Output: 2
Explanation: n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.

Example 3:
Input: nums = [9,6,4,2,3,5,7,0,1]
Output: 8
Explanation: n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.

Example 4:
Input: nums = [0]
Output: 1
Explanation: n = 1 since there is 1 number, so all numbers are in the range [0,1]. 1 is the missing number in the range since it does not appear in nums.

Constraints:
n == nums.length
1 <= n <= 104
0 <= nums[i] <= n
All the numbers of nums are unique.
*/

class Solution
{
public:
  int missingNumber(vector<int> &nums)
  {
    unordered_map<int, int> m;
    for (auto &i : nums)
      m[i]++;
    for (int i = 0; i <= nums.size(); ++i)
    {
      if (m[i] == 0)
        return i;
    }
    return -1;
  }
};
